In: Statistics and Probability
(1 point) Using the Standard Normal Table found in your textbook, find the z-scores such that:
(a) The area under the standard normal curve to its left is 0.5
z =
(b) The area under the standard normal curve to its left is 0.9826
z =
(c) The area under the standard normal curve to its right is 0.1423
z =
(d) The area under the standard normal curve to its right is 0.9394
z =
(a) The area under the standard normal curve to its left is 0.5
P(Z<x) = 0.5
From z table P = 0.5 obtained at z = 0
So Z score = 0
(b) The area under the standard normal curve to its left is 0.9826
P(Z<x) = 0.9826
From Z table p = 0.9826 Obtained z = 2.111
Z score = 2.111
(c) The area under the standard normal curve to its right is 0.1423
P(Z>x) = 0.1423
P(Z>x) = 1- P(Z<x) = 0.1423
1- P(Z<x) = 0.1423
P(Z<x) = 1-0.1423
P(Z<x) = 0.8577
From z table p = 0.8577 obtained at z = 1.07
Z score = 1.07
(d) The area under the standard normal curve to its right is 0.9394
P(Z>x) = 0.9394
P(Z>x) = 1- P(Z<x) = 0.9394
1- P(Z<x) = 0.9394
P(Z<x) = 1-0.9394
P(Z<x) = 0.0606
From z table p = 0.0606 obtained at z = -1.55
Z score = -1.55